Find the unit vector perpendicular to each of the vectors $6^i + 2^j + 3^k$ and $3^i - 2^k$ .

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Solution

$\begin{matrix} V e c t o r p e r p e n d i c u l a r t o 6^i + 2^j + 3^k \to c = (6^i + 2^j + 3^k) \times (3^i - 2^k) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 6 & 2 & 3 3 & 0 & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (- 4) -^j (- 12 - 9) +^k (0 - 6) = - 4^i + 21^j - 6^k I m p l i e s,^c = \frac{- 4^i + 2^j - 6^k}{\sqrt{16 + {(21)}^{2} + 36}} = \frac{- 4^i + 2^j - 6^k}{\sqrt{493}} \end{matrix}$

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Similar questions

\to a + \to b

\to a - \to b

\to a = 3^i + 2^j + 2^k

\to b =^i + 2^j - 2^k

\to a + \to b

\to a - \to b

\to a = 3^i + 2^j + 2^k

\to b =^i + 2^j - 2^k

Find a unit vector perpendicular to each of the vectors a + b and a - b, where $a = 3^i + 2^j + 2^k$ and $b =^i + 2^j - 2^k$

\to a + \to b

\to a - \to b

\to a = 3^i + 2^j + 2^k

\to b =^i + 2^j - 2^k

3^i -^j + 2^k

\to B = 2^i - 2^j + 4^k

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